\(\int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\) [1010]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {5 (7 A+B) \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {a (5 A-B)}{64 d (a+a \sin (c+d x))^2}-\frac {5 A}{32 d (a+a \sin (c+d x))} \]

[Out]

5/128*(7*A+B)*arctanh(sin(d*x+c))/a/d+1/96*a^2*(A+B)/d/(a-a*sin(d*x+c))^3+1/128*a*(5*A+3*B)/d/(a-a*sin(d*x+c))
^2+5/128*(3*A+B)/d/(a-a*sin(d*x+c))-1/64*a^3*(A-B)/d/(a+a*sin(d*x+c))^4-1/48*a^2*(2*A-B)/d/(a+a*sin(d*x+c))^3-
1/64*a*(5*A-B)/d/(a+a*sin(d*x+c))^2-5/32*A/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2915, 78, 212} \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {a^3 (A-B)}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}-\frac {a^2 (2 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {5 (7 A+B) \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}-\frac {a (5 A-B)}{64 d (a \sin (c+d x)+a)^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {5 A}{32 d (a \sin (c+d x)+a)} \]

[In]

Int[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

(5*(7*A + B)*ArcTanh[Sin[c + d*x]])/(128*a*d) + (a^2*(A + B))/(96*d*(a - a*Sin[c + d*x])^3) + (a*(5*A + 3*B))/
(128*d*(a - a*Sin[c + d*x])^2) + (5*(3*A + B))/(128*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(64*d*(a + a*Sin[c
+ d*x])^4) - (a^2*(2*A - B))/(48*d*(a + a*Sin[c + d*x])^3) - (a*(5*A - B))/(64*d*(a + a*Sin[c + d*x])^2) - (5
*A)/(32*d*(a + a*Sin[c + d*x]))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {a^7 \text {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {A+B}{32 a^5 (a-x)^4}+\frac {5 A+3 B}{64 a^6 (a-x)^3}+\frac {5 (3 A+B)}{128 a^7 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^5}+\frac {2 A-B}{16 a^5 (a+x)^4}+\frac {5 A-B}{32 a^6 (a+x)^3}+\frac {5 A}{32 a^7 (a+x)^2}+\frac {5 (7 A+B)}{128 a^7 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {a (5 A-B)}{64 d (a+a \sin (c+d x))^2}-\frac {5 A}{32 d (a+a \sin (c+d x))}+\frac {(5 (7 A+B)) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 d} \\ & = \frac {5 (7 A+B) \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {a (5 A-B)}{64 d (a+a \sin (c+d x))^2}-\frac {5 A}{32 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {15 (7 A+B) \text {arctanh}(\sin (c+d x))+\frac {48 (A-B)-33 (7 A+B) \sin (c+d x)-33 (7 A+B) \sin ^2(c+d x)+40 (7 A+B) \sin ^3(c+d x)+40 (7 A+B) \sin ^4(c+d x)-15 (7 A+B) \sin ^5(c+d x)-15 (7 A+B) \sin ^6(c+d x)}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^4}}{384 a d} \]

[In]

Integrate[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

(15*(7*A + B)*ArcTanh[Sin[c + d*x]] + (48*(A - B) - 33*(7*A + B)*Sin[c + d*x] - 33*(7*A + B)*Sin[c + d*x]^2 +
40*(7*A + B)*Sin[c + d*x]^3 + 40*(7*A + B)*Sin[c + d*x]^4 - 15*(7*A + B)*Sin[c + d*x]^5 - 15*(7*A + B)*Sin[c +
d*x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(384*a*d)

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\left (-\frac {35 A}{256}-\frac {5 B}{256}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {5 A}{64}-\frac {3 B}{64}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{32}+\frac {B}{32}}{3 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {\frac {15 A}{128}+\frac {5 B}{128}}{\sin \left (d x +c \right )-1}-\frac {5 A}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{16}-\frac {B}{16}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {A}{8}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}-\frac {B}{32}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {35 A}{256}+\frac {5 B}{256}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(170\)
default	\(\frac {\left (-\frac {35 A}{256}-\frac {5 B}{256}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {5 A}{64}-\frac {3 B}{64}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{32}+\frac {B}{32}}{3 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {\frac {15 A}{128}+\frac {5 B}{128}}{\sin \left (d x +c \right )-1}-\frac {5 A}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{16}-\frac {B}{16}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {A}{8}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}-\frac {B}{32}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {35 A}{256}+\frac {5 B}{256}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(170\)
parallelrisch	\(\frac {-525 \left (A +\frac {B}{7}\right ) \left (4+\frac {\sin \left (7 d x +7 c \right )}{5}+\sin \left (5 d x +5 c \right )+\frac {9 \sin \left (3 d x +3 c \right )}{5}+\sin \left (d x +c \right )+\frac {2 \cos \left (6 d x +6 c \right )}{5}+\frac {12 \cos \left (4 d x +4 c \right )}{5}+6 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+525 \left (A +\frac {B}{7}\right ) \left (4+\frac {\sin \left (7 d x +7 c \right )}{5}+\sin \left (5 d x +5 c \right )+\frac {9 \sin \left (3 d x +3 c \right )}{5}+\sin \left (d x +c \right )+\frac {2 \cos \left (6 d x +6 c \right )}{5}+\frac {12 \cos \left (4 d x +4 c \right )}{5}+6 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-8512 A -1216 B \right ) \cos \left (2 d x +2 c \right )+\left (-3752 A -536 B \right ) \cos \left (4 d x +4 c \right )+\left (-672 A -96 B \right ) \cos \left (6 d x +6 c \right )+\left (301 A +43 B \right ) \sin \left (3 d x +3 c \right )+\left (-735 A -105 B \right ) \sin \left (5 d x +5 c \right )+\left (-231 A -33 B \right ) \sin \left (7 d x +7 c \right )+\left (4389 A +627 B \right ) \sin \left (d x +c \right )-4920 A +2808 B}{384 a d \left (20+\sin \left (7 d x +7 c \right )+5 \sin \left (5 d x +5 c \right )+9 \sin \left (3 d x +3 c \right )+5 \sin \left (d x +c \right )+2 \cos \left (6 d x +6 c \right )+12 \cos \left (4 d x +4 c \right )+30 \cos \left (2 d x +2 c \right )\right )}\)	\(390\)
norman	\(\frac {\frac {\left (257 A +311 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}+\frac {\left (257 A +311 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}+\frac {5 \left (337 A +103 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}+\frac {5 \left (337 A +103 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}-\frac {\left (11 A -163 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}-\frac {\left (11 A -163 B \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}+\frac {\left (29 A +59 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a d}+\frac {\left (29 A +59 B \right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a d}+\frac {5 \left (5 A +19 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}+\frac {\left (93 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}+\frac {\left (93 A -5 B \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a d}+\frac {\left (41 A +719 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 a d}+\frac {\left (41 A +719 B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 a d}+\frac {\left (1363 A -299 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}+\frac {\left (1363 A -299 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {5 \left (7 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}+\frac {5 \left (7 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\)	\(483\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (105 A +15 B +170 i B \,{\mathrm e}^{9 i \left (d x +c \right )}+1190 i A \,{\mathrm e}^{9 i \left (d x +c \right )}+30 i B \,{\mathrm e}^{11 i \left (d x +c \right )}+210 i A \,{\mathrm e}^{11 i \left (d x +c \right )}+791 A \,{\mathrm e}^{4 i \left (d x +c \right )}+113 B \,{\mathrm e}^{4 i \left (d x +c \right )}+490 A \,{\mathrm e}^{2 i \left (d x +c \right )}+70 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1190 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-170 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-210 i A \,{\mathrm e}^{i \left (d x +c \right )}-30 i B \,{\mathrm e}^{i \left (d x +c \right )}+791 A \,{\mathrm e}^{8 i \left (d x +c \right )}+113 B \,{\mathrm e}^{8 i \left (d x +c \right )}+300 A \,{\mathrm e}^{6 i \left (d x +c \right )}-3468 B \,{\mathrm e}^{6 i \left (d x +c \right )}+396 i B \,{\mathrm e}^{7 i \left (d x +c \right )}-2772 i A \,{\mathrm e}^{5 i \left (d x +c \right )}-396 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+2772 i A \,{\mathrm e}^{7 i \left (d x +c \right )}+105 A \,{\mathrm e}^{12 i \left (d x +c \right )}+15 B \,{\mathrm e}^{12 i \left (d x +c \right )}+490 A \,{\mathrm e}^{10 i \left (d x +c \right )}+70 B \,{\mathrm e}^{10 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{128 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{128 a d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{128 a d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{128 a d}\)	\(487\)

[In]

int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*((-35/256*A-5/256*B)*ln(sin(d*x+c)-1)-1/2*(-5/64*A-3/64*B)/(sin(d*x+c)-1)^2-1/3*(1/32*A+1/32*B)/(sin(d*x
+c)-1)^3-(15/128*A+5/128*B)/(sin(d*x+c)-1)-5/32*A/(1+sin(d*x+c))-1/4*(1/16*A-1/16*B)/(1+sin(d*x+c))^4-1/3*(1/8
*A-1/16*B)/(1+sin(d*x+c))^3-1/2*(5/32*A-1/32*B)/(1+sin(d*x+c))^2+(35/256*A+5/256*B)*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {30 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} - 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} + 56 \, A + 8 \, B\right )} \sin \left (d x + c\right ) - 16 \, A - 112 \, B}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/768*(30*(7*A + B)*cos(d*x + c)^6 - 10*(7*A + B)*cos(d*x + c)^4 - 4*(7*A + B)*cos(d*x + c)^2 - 15*((7*A + B)
*cos(d*x + c)^6*sin(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 15*((7*A + B)*cos(d*x + c)^6*
sin(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(15*(7*A + B)*cos(d*x + c)^4 + 10*(7*A + B
)*cos(d*x + c)^2 + 56*A + 8*B)*sin(d*x + c) - 16*A - 112*B)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c
)^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{6} + 15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{5} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{4} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{3} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{2} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right ) - 48 \, A + 48 \, B\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{768 \, d} \]

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/768*(15*(7*A + B)*log(sin(d*x + c) + 1)/a - 15*(7*A + B)*log(sin(d*x + c) - 1)/a - 2*(15*(7*A + B)*sin(d*x +
c)^6 + 15*(7*A + B)*sin(d*x + c)^5 - 40*(7*A + B)*sin(d*x + c)^4 - 40*(7*A + B)*sin(d*x + c)^3 + 33*(7*A + B)
*sin(d*x + c)^2 + 33*(7*A + B)*sin(d*x + c) - 48*A + 48*B)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x
+ c)^5 - 3*a*sin(d*x + c)^4 + 3*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a))/d

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {60 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (385 \, A \sin \left (d x + c\right )^{3} + 55 \, B \sin \left (d x + c\right )^{3} - 1335 \, A \sin \left (d x + c\right )^{2} - 225 \, B \sin \left (d x + c\right )^{2} + 1575 \, A \sin \left (d x + c\right ) + 321 \, B \sin \left (d x + c\right ) - 641 \, A - 167 \, B\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {875 \, A \sin \left (d x + c\right )^{4} + 125 \, B \sin \left (d x + c\right )^{4} + 3980 \, A \sin \left (d x + c\right )^{3} + 500 \, B \sin \left (d x + c\right )^{3} + 6930 \, A \sin \left (d x + c\right )^{2} + 702 \, B \sin \left (d x + c\right )^{2} + 5548 \, A \sin \left (d x + c\right ) + 340 \, B \sin \left (d x + c\right ) + 1771 \, A - 35 \, B}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3072*(60*(7*A + B)*log(abs(sin(d*x + c) + 1))/a - 60*(7*A + B)*log(abs(sin(d*x + c) - 1))/a + 2*(385*A*sin(d
*x + c)^3 + 55*B*sin(d*x + c)^3 - 1335*A*sin(d*x + c)^2 - 225*B*sin(d*x + c)^2 + 1575*A*sin(d*x + c) + 321*B*s
in(d*x + c) - 641*A - 167*B)/(a*(sin(d*x + c) - 1)^3) - (875*A*sin(d*x + c)^4 + 125*B*sin(d*x + c)^4 + 3980*A*
sin(d*x + c)^3 + 500*B*sin(d*x + c)^3 + 6930*A*sin(d*x + c)^2 + 702*B*sin(d*x + c)^2 + 5548*A*sin(d*x + c) + 3
40*B*sin(d*x + c) + 1771*A - 35*B)/(a*(sin(d*x + c) + 1)^4))/d

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^4+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,\sin \left (c+d\,x\right )-\frac {A}{8}+\frac {B}{8}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^7-a\,{\sin \left (c+d\,x\right )}^6+3\,a\,{\sin \left (c+d\,x\right )}^5+3\,a\,{\sin \left (c+d\,x\right )}^4-3\,a\,{\sin \left (c+d\,x\right )}^3-3\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (7\,A+B\right )}{128\,a\,d} \]

[In]

int((A + B*sin(c + d*x))/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

(B/8 - A/8 + sin(c + d*x)*((77*A)/128 + (11*B)/128) - sin(c + d*x)^3*((35*A)/48 + (5*B)/48) - sin(c + d*x)^4*(
(35*A)/48 + (5*B)/48) + sin(c + d*x)^5*((35*A)/128 + (5*B)/128) + sin(c + d*x)^6*((35*A)/128 + (5*B)/128) + si
n(c + d*x)^2*((77*A)/128 + (11*B)/128))/(d*(a + a*sin(c + d*x) - 3*a*sin(c + d*x)^2 - 3*a*sin(c + d*x)^3 + 3*a
*sin(c + d*x)^4 + 3*a*sin(c + d*x)^5 - a*sin(c + d*x)^6 - a*sin(c + d*x)^7)) + (5*atanh(sin(c + d*x))*(7*A + B
))/(128*a*d)